The problem involves a boy on roller blades who throws a jug of water, and the objective is to determine the velocity gained by the boy after the throw. The context is rooted in the principles of conservation of momentum.
Some guidance has been offered regarding the conservation of momentum, with participants attempting to outline the equations involved. However, there is a lack of explicit consensus on the final interpretation of the results, and some participants question the completeness of the provided reasoning.
There are indications that the discussion is constrained by the forum's rules against providing complete solutions, which has led to some reiteration of the problem setup and the need for clarity in the approach taken.
you are not supposed to solve the whole problem. this is against pf rulesmido808 said:so you apply the formula P(before)=P(after). (And by the way the momentum before is 0 as the intial velocities of the boy and the jug are 0 as P(before)= m(boy)xv(boyintial)+m(jug)xv(jug int.))Hope u understood until now.
p(after)=m(boy)xv(boy final)+m(jug)xv(jug)
P(after)=38kg(v)+7.9kg(14m/s)
P(after)=38v+110.6
P(before)=p(after)
0=38v+110.6
38v=-110.6
V=-2.9
So velocity of the boy is 2.9 and its direction is in the opposite direction that he threw the jug in.
Thanks for reading in my answer and i hope u understand and everbody as well since i am a new user.